University Quantum Mechanics II · Time-Dependent Perturbation Theory · 12.3

First-order amplitudes and a perturbation turned on at t = 0

This is where the course first computes a transition probability instead of a level shift. The integral takes one line. The work is in reading its answer honestly: which part is a real transition, which part is the state rearranging itself around a new Hamiltonian, and why a spread of final energies costs no energy at all.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Quantum Mechanics I evolved states only under a Hamiltonian that never changed. Here H′ switches on at t = 0, and the exact coefficient equations couple every amplitude to every other. First order makes one bold replacement: while the system has barely left |n⟩, set cₙ = 1 and every other coefficient to zero on the right-hand side.

What survives is a single integral, cₘ(t) = −(i/ħ)∫₀ᵗ H′ₘₙ(t′)e^{(iωₘₙt′)}dt′: the Fourier component of the perturbation's matrix element, over the time it has acted, at the Bohr frequency of the pair. For a constant H′ switched on suddenly it gives Pₘ = 4|H′ₘₙ|²sin²(ωₘₙt/2)/(ħωₘₙ)², and that answer has to be read with care. It oscillates rather than accumulates.

Half of it is the stationary admixture Unit 6 already found; the other half is a transient that exists only because the switch was sudden, and a switch slower than 1/|ωₘₙ| removes it. Read at fixed t across final levels, it is a peak whose height grows as t² while its width shrinks as 1/t: a property of Fourier integrals over a finite window, paid for by ⟨H′⟩, with energy conserved throughout. The model stops when the initial state is noticeably depleted, at t ≈ ħ/|H′ₘₙ| for a degenerate pair, after long times when the neglected second-order level shifts put its phases out of step, and when the final levels crowd into a continuum and the t² peak must be summed into a rate.

Simple definition: The first-order amplitude cₘ(t) is the amplitude for reaching |m⟩ when the perturbation may act only once on the initial state |n⟩, at any moment t′ between 0 and t, each moment weighted by the phase e^{(iωₘₙt′)}.
Example: A 5.0 mT field stepped on along z couples hydrogen's F = 0 ground state to F = 1, M = 0 with H′ₘₙ = 0.290 μeV against a 5.874 μeV splitting, so P swings between 0 and 4(0.0494)² = 0.98% and returns to zero every 0.704 ns.

First-order amplitudecₘ(t) = −(i/ħ) ∫₀ᵗ H′ₘₙ(t′) e^{(iωₘₙt′)} dt′, with ωₘₙ = (Eₘ − Eₙ)/ħ

Derived by setting cₖ = δₖₙ on the right of the exact coefficient equations: the first Dyson term. It is the Fourier transform of H′ₘₙ over [0, t] at the Bohr frequency.

cₘ dimensionless (m ≠ n); H′ₘₙ = ⟨m|H′|n⟩ in J; ħ in J s; t in s; ωₘₙ in rad s⁻¹; starting from cₖ(0) = δₖₙ

Constant H′ switched on at t = 0: amplitudecₘ(t) = (H′ₘₙ/ħωₘₙ)(1 − e^(iωₘₙt))

Derived. The −e^(iωₘₙt) term is Unit 6's admixture of |m⟩ in the perturbed |n′⟩; the constant term is the transient the sudden switch leaves, equal to ⟨m′|n⟩.

H′ zero before t = 0 and constant after; H′ₘₙ/ħωₘₙ is dimensionless, a ratio of two energies

Constant H′ switched on at t = 0: probabilityPₘ(t) = 4|H′ₘₙ|² sin²(ωₘₙt/2) / (ħωₘₙ)²

Derived. As ωₘₙ → 0 it becomes |H′ₘₙ|²t²/ħ². Off resonance it returns to zero every 2π/|ωₘₙ| with H′ still on, so nothing accumulates.

Dimensionless. Swings at ωₘₙ between 0 and 4|H′ₘₙ|²/(ħωₘₙ)², averaging twice the admixture |H′ₘₙ|²/(ħωₘₙ)²

The same result read across final levelsPₘ(ωₘₙ) = (|H′ₘₙ|t/ħ)² [sin(ωₘₙt/2) / (ωₘₙt/2)]²

Derived by rewriting the step result. The height grows as t² while the zeros close in as 1/t, so the area grows only as t: the seed of Fermi's golden rule.

t held fixed; the bracket is dimensionless, 1 at ωₘₙ = 0, first zero at ωₘₙ = ±2π/t, at most 0.047 in the side lobes

Integration by parts for any switch-oncₘ(t) = −H′ₘₙ(t)e^(iωₘₙt)/ħωₘₙ + (1/ħωₘₙ) ∫₀ᵗ (dH′ₘₙ/dt′) e^{(iωₘₙt′)} dt′

Derived. The first term follows H′ instant by instant; the second is the switch's own Fourier component at ωₘₙ, cut to 1/√(1 + ωₘₙ²τ²) of a step's by an exponential rise of time constant τ.

H′ₘₙ(t) rises from zero at t = 0; a sudden step puts H′ₘₙδ(t′) into the derivative; needs ωₘₙ ≠ 0

When the first-order answer holdsΣₘ Pₘ ≪ 1: |H′ₘₙ| ≪ ħ|ωₘₙ| off resonance, and |H′ₘₙ|t/ħ ≪ 1 once |ωₘₙ|t ≲ 1

Derived from the step result's ceiling and its small-ω limit. Against QM I's exact two-state answer the first-order ceiling is high by the factor 1 + 4|H′ₘₙ|²/(ħωₘₙ)².

Both compare the coupling, in J, with an energy: the level spacing ħ|ωₘₙ|, or ħ/t

Freeze the right-hand side at the initial state

The exact equations of this unit, iħ dcₘ/dt = Σₖ H′ₘₖ(t)e^(iωₘₖt)cₖ(t), couple every amplitude to every other, because population arriving in |m⟩ can at once feed |k⟩. While the system has hardly left |n⟩, the right-hand side is dominated by the one term with cₙ ≈ 1. Put cₖ = δₖₙ there and integrate from 0: cₘ(t) = −(i/ħ)∫₀ᵗ H′ₘₙ(t′)e^{(iωₘₙt′)}dt′, the first term of the Dyson series, whose single time needs no ordering. The dimensions check at once: H′ₘₙt/ħ is J s over J s. The m = n equation is a limit check. For constant H′ it gives cₙ ≈ 1 − iH′ₙₙt/ħ, the first two terms of e^{(−iH′ₙₙt/ħ)}, so Unit 6's first-order shift E⁽¹⁾ = H′ₙₙ appears as a phase and a diagonal H′ causes no transition. The reading to carry forward is that cₘ is the Fourier transform of H′ₘₙ(t), cut off at 0 and t, evaluated at the pair's Bohr frequency. A pair responds only to the part of the perturbation's time dependence that matches its own level spacing.

A step in H′ makes the probability oscillate, not accumulate

Let H′ be zero before t = 0 and constant after. The integral is elementary: cₘ = −(i/ħ)H′ₘₙ(e^(iωₘₙt) − 1)/(iωₘₙ) = (H′ₘₙ/ħωₘₙ)(1 − e^(iωₘₙt)), and |1 − e^(iθ)|² = 4sin²(θ/2) gives Pₘ = 4|H′ₘₙ|²sin²(ωₘₙt/2)/(ħωₘₙ)². As ωₘₙ → 0 it becomes |H′ₘₙ|²t²/ħ², the quadratic start of a degenerate pair. Off resonance P returns to exactly zero at t = 2π/|ωₘₙ| with H′ still on. Take hydrogen in its F = 0 ground state and step on 5.0 mT along z. The splitting to F = 1, M = 0 is measured as 1420.405 751 768 MHz, or 5.874 μeV; Unit 9's (4/3)gₚ(mₑ/mₚ)α⁴mₑc² = 5.877 μeV, an α⁴ effect, is quoted. Using ⟨1,0|S_z|0,0⟩ = ħ/2 = −⟨1,0|I_z|0,0⟩, the coupling is H′ₘₙ = (gₑμ_B + gₚμ_N)B/2 = 0.2902 μeV with measured g-factors, so H′ₘₙ/ħωₘₙ = 0.0494 and P swings between 0 and 0.98% every 0.704 ns.

Half of it is Unit 6's admixture, half is the switch

Split the step amplitude as cₘ = −(H′ₘₙ/ħωₘₙ)e^(iωₘₙt) + H′ₘₙ/ħωₘₙ. Unit 6 built the perturbed stationary state |n′⟩ = |n⟩ + Σₘ|m⟩H′ₘₙ/(Eₙ − Eₘ), which evolves as e^{(−iEₙt/ħ)}. Its |m⟩ amplitude, multiplied by e^(iEₘt/ħ) to reach this picture, is −(H′ₘₙ/ħωₘₙ)e^(iωₘₙt): exactly the first piece. That part is not a transition. It is |m⟩ belonging to the new stationary state, and a slow switch-off hands it back to |n⟩. The constant piece is ⟨m′|n⟩ = H′ₘₙ/(Eₘ − Eₙ). Just after a sudden switch the system is still in |n⟩, which is not an eigenstate of H₀ + H′ and so has a small component along the perturbed |m′⟩; that component carries the phase e^{(−iEₘt/ħ)}, so it is constant here. The two pieces are equal in size and beat at ωₘₙ. For the hydrogen pair the admixture is 0.0494² = 0.244%, P averages twice that and peaks at four times it, and only the transient's 0.244% survives as population in F = 1 if the field is later removed slowly.

Integrate by parts to see how slow is slow

For any H′(t) that starts from zero, integrate the first-order integral by parts: cₘ(t) = −H′ₘₙ(t)e^(iωₘₙt)/ħωₘₙ + (1/ħωₘₙ)∫₀ᵗ(dH′ₘₙ/dt′)e^{(iωₘₙt′)}dt′. The first term is the admixture, following H′ instant by instant. The second is the transient: the Fourier component, at the Bohr frequency, of the rate at which the switch changes H′. A step has dH′/dt = H′δ(t′) and returns H′ₘₙ/ħωₘₙ. An exponential rise H′(1 − e^(−t/τ)) gives, for t ≫ τ, the step's transient times 1/(1 − iωₘₙτ), smaller by 1/√(1 + ωₘₙ²τ²). A switch is sudden if τ ≪ 1/|ωₘₙ| and adiabatic if τ ≫ 1/|ωₘₙ|. For the hyperfine pair 1/ω = 0.112 ns: a rise with τ = 0.1 ns keeps 75% of the transient, but a coil with τ = 1 μs has ωτ = 8.9 × 10³ and leaves P = 0.244% with a ripple of ±5 × 10⁻⁷. For 1s → 2p the Lyman-α spacing, measured as 10.2 eV at 121.57 nm and of order α²mₑc², gives 1/ω = 0.065 fs. No laboratory field switches that fast, so a static field only polarises the atom.

Across final levels, the peak narrows as it grows

Now fix t and ask how P depends on ωₘₙ, as if many final levels, or one whose spacing can be tuned, were waiting. Writing Pₘ = (|H′ₘₙ|t/ħ)²[sin(ωₘₙt/2)/(ωₘₙt/2)]² shows the shape: a central peak of height |H′ₘₙ|²t²/ħ², first zeros at ωₘₙ = ±2π/t, and side lobes no higher than 4.7% of the peak. Double t and the peak grows fourfold while its width halves. The area, 2π|H′ₘₙ|²t/ħ², grows only as t, the fact Fermi's golden rule will turn into a rate. In energy the first zeros sit at |Eₘ − Eₙ| = h/t: 4.14 μeV after 1 ns, 4.14 neV after 1 μs. Energy conservation is not suspended. Once H′ is on, the Hamiltonian H₀ + H′ does not depend on time, so ⟨H₀ + H′⟩ is exactly conserved. To second order ⟨H₀⟩ rises by ħωₘₙPₘ while ⟨H′⟩ falls from H′ₙₙ by 4|H′ₘₙ|²sin²(ωₘₙt/2)/ħωₘₙ, the same amount; at the hydrogen maximum each moves by 57 neV. The width is what any signal lasting a time t has: a Fourier spread of about 2π/t.

How long the first-order answer lasts

The approximation replaced cₙ by 1, so it holds while Σₘ Pₘ ≪ 1. Off resonance the step result never exceeds 4|H′ₘₙ|²/(ħωₘₙ)², so |H′ₘₙ| ≪ ħ|ωₘₙ| keeps it small for all time; near resonance P ≈ |H′ₘₙ|²t²/ħ² demands t ≪ ħ/|H′ₘₙ|. The hydrogen pair is an exact two-state problem, since a field along z connects F = 0 only to F = 1, M = 0, so QM I's static-coupling solution is the benchmark: P = [4|V|²/(ħΩ)²]sin²(Ωt/2), with ħΩ = √((ħω)² + 4|V|²). At 5.0 mT it peaks at 0.967% against first order's 0.976%, and returns to zero every 0.701 ns against 0.704 ns. The ceiling is good to 1%, but first order's period is 0.49% too long, because it omits the second-order shifts that push the two levels apart. That phase error grows linearly. After π/(Ω − ω) = 72 ns, about 100 periods, the first-order curve peaks where the exact one vanishes. Small P is necessary, not sufficient: at late times the phase must be right too, and the next lessons measure that failure against the exact Rabi solution.

Change one variable at a time

Make the relationship visible.

Interactive model

Time since switch-on t60 ns

Linear ramp duration T0 ns

At T = 0, raise t from 40 to 80 ns: the peak grows fourfold, its zeros close in from ±25 to ±12.5 MHz, and the wings swing between zero and four times the dashed admixture. Now raise T: wherever νT passes about 1, the dips fill in and the solid curve settles onto the dashed one.

PEAK P AT ν = 00.0355

FIRST ZEROS ±1/(t − T/2)16.7 MHz

ENERGY WINDOW h/(t − T/2)68.9 neV

RAMP ÷ STEP TRANSIENT, 30 MHz1.000

Live interpretationPEAK P AT ν = 0: 0.0355. FIRST ZEROS ±1/(t − T/2): 16.7 MHz. ENERGY WINDOW h/(t − T/2): 68.9 neV. RAMP ÷ STEP TRANSIENT, 30 MHz: 1.000

Catch the common trap

Explain before calculating.

Misconception check

The tempting reading of the central peak's width is that, for a short time, the system may borrow an energy of order h/t and so visit levels that energy conservation would forbid, with ΔE Δt ≥ ħ/2 as the licence. Nothing is borrowed. After t = 0 the Hamiltonian H₀ + H′ is constant, so ⟨H₀ + H′⟩ is exactly conserved at every instant, however large P becomes. What spreads is H₀, which is no longer the Hamiltonian: finding the system in |m⟩ raises ⟨H₀⟩ by ħωₘₙPₘ, and the coupling energy ⟨H′⟩ falls by precisely that amount. The only energy exchanged with the outside is paid at t = 0, when whatever switches H′ on does work H′ₙₙ on the state. The width 4π/t between the first zeros is a property of a Fourier integral over a window of length t, the same trade-off that limits any frequency measurement lasting t to a resolution of about 1/t.

A constant perturbation with |H′ₘₙ| = 0.10ħ|ωₘₙ| acts on a system that starts in |n⟩. Stepped on at t = 0, it makes Pₘ oscillate between 0 and 0.040. What does Pₘ do at late times if the same perturbation is instead switched on gradually, as H′(1 − e^(−t/τ)) with τ = 50/|ωₘₙ|?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyAn electron spin in a field B₀ = 0.350 T along z starts spin-down, its lower state. At t = 0 a transverse field Bₓ = 2.1 mT is switched on suddenly. With gₑμ_B/h = 28.025 GHz T⁻¹, find the first-order spin-flip probability at t = 30.0 ps, its largest value and when that is first reached, and check the ceiling against QM I's exact precession.

The pair: ħωₘₙ = gₑμ_B B₀, so ν = 28.025 × 0.350 = 9.809 GHz and ωₘₙ = 2πν = 6.163 × 10¹⁰ s⁻¹. The coupling H′ = gₑμ_B BₓSₓ/ħ has ⟨↑|Sₓ|↓⟩ = ħ/2, so H′ₘₙ = gₑμ_B Bₓ/2 and H′ₘₙ/ħωₘₙ = Bₓ/2B₀ = 0.00300, a dimensionless ratio of fields.
Step result: P(t) = 4(0.00300)² sin²(ωₘₙt/2) = 3.60 × 10⁻⁵ sin²(πνt). At t = 30.0 ps, πνt = 0.9245 rad and sin² = 0.637, so P = 2.29 × 10⁻⁵.
The ceiling 3.60 × 10⁻⁵ is first reached when πνt = π/2, at t = 1/(2ν) = 51.0 ps, and P is back to zero at 102 ps although Bₓ stays on.
Exact check: the spin precesses about the tilted total field, at angle θ with tan θ = Bₓ/B₀ = 0.00600, and its largest flip probability is sin²θ. First order's 4(Bₓ/2B₀)² equals tan²θ, which exceeds sin²θ only by the factor 1 + tan²θ = 1.00004, so both give 3.60 × 10⁻⁵.

AnswerP(30.0 ps) = 2.29 × 10⁻⁵; the largest value, 3.60 × 10⁻⁵, is first reached at 51.0 ps, and the exact precession agrees to 0.004%.

MediumIn a double quantum dot an electron starts in the left-dot level |n⟩. At t = 0 a gate pulse switches on a tunnel coupling V = 0.050 μeV to the right-dot level |m⟩, fast enough to treat as a step. Find Pₘ at t = 2.0 ns and at 4.0 ns for detunings Eₘ − Eₙ = 0 and 1.00 μeV, and explain why waiting longer lowers one of them.

Scales: ħ = 0.6582 μeV ns, so for ε = 1.00 μeV, ωₘₙ = 1.519 × 10⁹ s⁻¹ and 1/ωₘₙ = 0.658 ns. A pulse that rises in much less than that is a step, so the step formula applies.
Resonant level: P = (Vt/ħ)². At 2.0 ns, Vt/ħ = 0.050 × 2.0/0.6582 = 0.1519 and P = 0.0231; doubling t doubles Vt/ħ to 0.3039 and quadruples P to 0.0923.
Level at 1.00 μeV: P = 4(V/ε)² sin²(ωₘₙt/2), with 4(V/ε)² = 0.0100. At 2.0 ns, ωₘₙt/2 = 1.519 rad, sin² = 0.997 and P = 0.00997. Equivalently, the resonant 0.0231 times the sinc² factor 0.432: the level lies inside the central peak, whose first zeros sit at h/t = 2.07 μeV.
At 4.0 ns, ωₘₙt/2 = 3.039 rad, just short of π, so sin² = 0.0106 and P = 1.06 × 10⁻⁴, 94 times smaller. The window h/t = 1.03 μeV has shrunk onto the level, which now sits almost at the first zero. So close to a zero P is fragile: an exponential rise with τ = 0.1 ns keeps 99% of the transient's size but turns its phase by arctan(ωₘₙτ) = 0.15 rad, and would raise this value to 3.1 × 10⁻⁴.
Exact check with QM I's two-state result, P = [4V²/(ε² + 4V²)]sin²(Ωt/2) with ħΩ = √(ε² + 4V²): 0.0229 and 0.0895 on resonance, and 0.00988 at 2.0 ns off it. At 4.0 ns the exact 7.6 × 10⁻⁵ lies 28% below first order: near a zero the absolute error, 3 × 10⁻⁵, is small, but the relative one is not.

AnswerResonant: 0.0231 at 2.0 ns and 0.0923 at 4.0 ns. Detuned by 1.00 μeV: 0.00997 at 2.0 ns but 1.06 × 10⁻⁴ at 4.0 ns, because the peak's width h/t has shrunk onto the level. Exact: 0.0229, 0.0895, 0.00988 and 7.6 × 10⁻⁵.

HardHydrogen atoms in the F = 0 hyperfine ground state (splitting measured as 1420.405 751 768 MHz) see a field along z rise to 5.0 mT, where H′ₘₙ = 0.2902 μeV couples them to F = 1, M = 0. The field is held, then removed linearly over 1 μs. Find the population left in F = 1 when the field is stepped on, ramped on linearly over T = 1.00 ns, or ramped on over T = 0.704 ns.

Integrate by parts over the whole history. Once the field is gone the admixture term −H′ₘₙ(t)e^(iωₘₙt)/ħωₘₙ is zero, so cₘ = (1/ħωₘₙ)∫(dH′ₘₙ/dt′)e^{(iωₘₙt′)}dt′: only the switching counts. The 1 μs removal contributes at most 2/(ωₘₙ × 1 μs) = 2 × 10⁻⁴ of a step's amplitude, negligible here.
Numbers: ωₘₙ = 2π × 1.4204 × 10⁹ Hz = 8.925 × 10⁹ s⁻¹, ħωₘₙ = 5.874 μeV, so a = H′ₘₙ/ħωₘₙ = 0.0494 and a² = 2.44 × 10⁻³.
Step on: dH′/dt = H′δ(t′), so cₘ = a and P = a² = 2.44 × 10⁻³, whatever the holding time. The slow removal hands the admixture back to F = 0 but carries the sudden switch's transient into F = 1.
Linear ramp: dH′/dt = H′/T while t′ runs from 0 to T, so cₘ = a(e^(iωₘₙT) − 1)/(iωₘₙT) and P = a²[sin(ωₘₙT/2)/(ωₘₙT/2)]². For T = 1.00 ns, ωₘₙT/2 = 4.462 rad and sin = −0.969, so the bracket is −0.217 and P = 2.44 × 10⁻³ × 0.0471 = 1.15 × 10⁻⁴. While the field is held, P swings only between 1.50 × 10⁻³ and 3.62 × 10⁻³, not 0 to 9.8 × 10⁻³.
For T = 0.704 ns = 1/ν, exactly one hyperfine period, ωₘₙT/2 = π and P = 0 at first order: a ramp lasting one Bohr period has no Fourier component at the Bohr frequency. As T → 0 the bracket tends to 1 and the step is recovered; for ωₘₙT ≫ 1 it falls below 2/(ωₘₙT), the adiabatic limit.
Check by integrating the exact two-state equations numerically: 2.42 × 10⁻³, 1.14 × 10⁻⁴ and about 1.5 × 10⁻⁸, within 1.2% of first order where that is nonzero; where it gives zero, the residue scales as a⁶ = 1.5 × 10⁻⁸, a third-order effect.

AnswerStepped on: 2.44 × 10⁻³ left in F = 1. Ramped over 1.00 ns: 1.15 × 10⁻⁴. Ramped over 0.704 ns: zero at first order (about 1.5 × 10⁻⁸ from the exact equations), because a ramp lasting one Bohr period carries no Fourier component at ωₘₙ.