University Quantum Mechanics II · Time-Dependent Perturbation Theory · 12.1
Coefficient equations for H₀ + H′(t)
Every rate in the rest of this course for absorption, stimulated emission, decay into a continuum and scattering is read off one set of equations. This topic writes them down with nothing thrown away, checks them against a problem already solved, and says exactly what their squared amplitudes are probabilities of.
Build the model
Connect the measurement to the mechanism.
Quantum Mechanics I evolved any state under a fixed Hamiltonian by expanding it on the eigenstates and attaching the phases e(−iEₙt/ħ). Atoms in fields, spins in pulses and anything being driven have a Hamiltonian H₀ + H′(t) that changes, and the move here is to keep the solved part exact while letting the rest act. Write the state on the eigenstates of H₀ with their phases pulled out, so that each amplitude cₙ(t) records only what H′ has done.
Because H₀|n⟩ = Eₙ|n⟩ exactly, the H₀ terms cancel, and one projection turns the Schrödinger equation into coupled first-order equations in which each cₘ is driven by the matrix elements H′ₘₙ(t), each carrying a phase that turns at the Bohr frequency ωₘₙ. Nothing has been assumed small: the equations are exact, Hermiticity keeps Σ|cₙ|² = 1, and for two levels they are the static pair of Quantum Mechanics I made time-dependent. What they cannot do is decide the physics for you.
The |cₙ|² are probabilities for the states of H₀, and become energy probabilities only once H′ is off. The rotating phases make distant levels hard to reach, which is why a model keeps only a few, but how many is a choice that has to be tested. And H′ must be a prescribed function of time, so a quantised field, and the spontaneous decay it causes, lies outside them.
- Simple definition
- The coefficient equations are the Schrödinger equation for H₀ + H′(t) rewritten, with no approximation, as coupled first-order equations for the amplitudes cₙ(t) of the unperturbed stationary states, each state keeping its own phase e(−iEₙt/ħ).
- Example
- A proton in B₀ = 50.0 μT, given a transverse B₁ = 30.0 μT at t = 0, obeys dc↓/dt = (iω₁/2)e(iω₀t)c↑ and its partner, with ω₀ = 1.34 × 10⁴ rad s⁻¹ and ω₁/2 = 4.01 × 10³ s⁻¹; solved exactly, |c↓|² never exceeds 30²/(50² + 30²) = 0.265.
Exact for any state, since the |n⟩ are complete. The phases carry all of H₀'s motion, so with H′ = 0 every cₙ is constant, and whatever changes a cₙ is H′.
H₀|n⟩ = Eₙ|n⟩ and ⟨m|n⟩ = δₘₙ; Eₙ in J, t in s, ħ in J s; the cₙ are dimensionless
One equation per state of H₀, with nothing assumed small. Diagonal elements only turn phases; off-diagonal ones move probability between the |n⟩.
H′ₘₙ(t) = ⟨m|H′(t)|n⟩ in J; ωₘₙ = (Eₘ − Eₙ)/ħ in rad s⁻¹; both sides are energies
Derived. A numerical solution whose norm drifts is wrong; one whose norm holds may still be wrong, because any Hermitian mistake keeps it too.
since H′ₙₘ = H′ₘₙ* and ωₙₘ = −ωₘₙ, the matrix H′ₘₙe(iωₘₙt) is Hermitian at every t
Derived: the general equations written out for n = 1, 2. The same pair holds for a step, a pulse or an oscillating drive; only the matrix elements change.
ċ = dc/dt in s⁻¹, H′₂₁ = H′₁₂*. For constant H′, aₙ = cₙe(−iEₙt/ħ) gives back Quantum Mechanics I's static pair
Derived here by eliminating c₁. For |V| ≪ ħω₀ the ceiling is about (2|V|/ħω₀)², the test for whether a distant level may be dropped.
H′₁₂ = V from t = 0, H′₁₁ = H′₂₂ = 0, c₁(0) = 1; Ω = √(ω₀² + 4|V|²/ħ²) in rad s⁻¹
Derived: with H′ = 0 every dcₙ/dt vanishes, and amplitudes are continuous across a sudden switch, so the values at τ become, and stay, the energy distribution.
τ in s is the time after which H′ = 0 for good. While H′ ≠ 0, |cₙ|² is the probability of the state |n⟩, not of an energy
Keep the solved motion in the phases
Quantum Mechanics I solved every fixed Hamiltonian the same way: expand on its eigenstates and attach e(−iEₙt/ħ) to each. With H = H₀ + H′(t) that recipe fails in general, because a Hamiltonian that changes has no fixed eigenstates to expand on. Dirac's move is to keep the old basis and the old phases anyway, writing |Ψ(t)⟩ = Σₙ cₙ(t)e(−iEₙt/ħ)|n⟩ with H₀|n⟩ = Eₙ|n⟩. With H′ = 0 this is the old solution with constant cₙ, so any change in a cₙ is caused by H′ alone, and the fast motion H₀ generates is carried in closed form. The two rates can differ enormously. In hydrogen the 2p phase turns relative to 1s at ω₂₁ = (3/8)α²mc²/ħ = 1.55 × 10¹⁶ rad s⁻¹, from Unit 2's energies, while a field of 1.0 × 10⁷ V m⁻¹ couples the pair through eE⟨210|z|100⟩ = eE(0.745a₀) = 3.94 × 10⁻⁴ eV, a rate of 5.99 × 10¹¹ s⁻¹, some 26 000 times slower. The basis must be complete, so for hydrogen the sum over n includes the continuum as an integral.
One cancellation and one projection
Substitute the expansion into iħ∂|Ψ⟩/∂t = (H₀ + H′)|Ψ⟩. The derivative acts on both cₙ and its phase, so the left side is Σₙ(iħ dcₙ/dt + Eₙcₙ)e(−iEₙt/ħ)|n⟩, and the right side is Σₙ cₙe(−iEₙt/ħ)(Eₙ + H′)|n⟩. The Eₙcₙ terms cancel exactly, and only because H₀|n⟩ = Eₙ|n⟩: this is where H₀ must already be solved. Apply ⟨m|, use ⟨m|n⟩ = δₘₙ and multiply by e(iEₘt/ħ): iħ dcₘ/dt = Σₙ H′ₘₙ(t)e(iωₘₙt)cₙ, with ωₘₙ = (Eₘ − Eₙ)/ħ. Nothing was approximated. There are as many equations as states, and they hold for any H′(t), including a step: the right side stays finite, so each cₙ is continuous across a sudden switch. Dimensions: ħ dcₘ/dt and H′ₘₙ are both energies. Limiting case: if H′ is constant and diagonal in |n⟩, then cₘ(t) = cₘ(0)e(−iH′ₘₘt/ħ). Unit 6's first-order shift H′ₘₘ, exact here because nothing sits off the diagonal, appears as a phase and moves no probability; only off-diagonal elements do that.
Why Σ|cₙ|² stays 1, and what that checks
Collect the right-hand side as a matrix Mₘₙ(t) = H′ₘₙ(t)e(iωₘₙt). H′ is Hermitian, so H′ₙₘ = H′ₘₙ*, and ωₙₘ = −ωₘₙ, so Mₙₘ = Mₘₙ*: M is Hermitian at every instant. Then iħ d(c†c)/dt = c†Mc − (Mc)†c = 0, and the total probability is 1 for all time whatever energy the source of H′ supplies; that energy redistributes population and never creates norm. The result is derived, and it is the first test of any numerical solution. Forward Euler replaces each step by 1 − iMΔt/ħ, which for two levels coupled only off the diagonal, by ħω₁/2, multiplies the norm by exactly 1 + (ω₁Δt/2)² per step: 10⁴ steps with ω₁Δt/2 = 0.01 inflate it to 2.72. For the proton of the Medium example, a 1 μs step lets Euler's norm grow by 0.72% over 1 ms, while fourth-order Runge-Kutta at the same step holds it to 4 parts in 10¹³. The test is one-sided, though. Drop the ½ from ω₁/2 and the matrix stays Hermitian and the norm exactly 1, yet for B₀ = 50.0 μT and B₁ = 20.0 μT the ceiling comes out 0.390 instead of 0.138.
Two levels: the static pair, made time-dependent
With levels |1⟩ and |2⟩ and ω₀ = (E₂ − E₁)/ħ, the equations are iħċ₁ = H′₁₁c₁ + H′₁₂e(−iω₀t)c₂ and iħċ₂ = H′₂₁e(iω₀t)c₁ + H′₂₂c₂. For a constant H′, the amplitudes aₙ = cₙe(−iEₙt/ħ) obey iħȧ = (H₀ + H′)a, the pair Quantum Mechanics I solved, so the new equations must reproduce the old answer. Take H′₁₁ = H′₂₂ = 0, H′₁₂ = V and c₁(0) = 1. Eliminating c₁ gives c̈₂ − iω₀ċ₂ + (|V|²/ħ²)c₂ = 0; putting c₂ = e(iω₀t/2)f leaves f̈ + (Ω/2)²f = 0 with Ω = √(ω₀² + 4|V|²/ħ²), and ċ₂(0) = −iV*/ħ gives |c₂|² = (4|V|²/ħ²Ω²)sin²(Ωt/2). It is dimensionless, vanishes as V → 0, and becomes a full exchange, sin²(|V|t/ħ), when ω₀ → 0. For a proton, with the measured γ/2π = 42.58 MHz T⁻¹, a 50.0 μT field gives ω₀/2π = 2.129 kHz, the signal a proton-precession magnetometer counts to read the Earth's field. A transverse B₁ supplies V = −ħγB₁/2, and |c↓|² is the spin precessing about the tilted total field.
What |cₙ(t)|² is a probability of
|cₙ(t)|² = |⟨n|Ψ(t)⟩|² is the probability that a measurement whose eigenstates are the |n⟩ returns n. While H′ acts, those are not the energy eigenstates. The energy is that of H₀ + H′, and for a constant H′ its eigenstates are fixed mixtures of |1⟩ and |2⟩ whose probabilities do not change after switch-on. In the figure's default, B₀ = 50.0 μT and B₁ = 30.0 μT tilt the total field 31.0° from z. The upper energy state, spin against that field, then holds probability sin²(15.5°) = 0.071 throughout, while |c↓|² swings between 0 and 0.265 at 2.483 kHz: that swing is Sz changing as the spin precesses, which an analyser along z would record. Switch B₁ off at τ and H₀ is the whole Hamiltonian again. Every dcₙ/dt is then zero, the amplitudes keep the values they had at τ, and |cₙ(τ)|² is the energy distribution from then on. A transition probability in this unit always means that: the probability of being found in |m⟩ once the perturbation is over.
Keeping a few levels is a model, and can be tested
The factor e(iωₘₙt) reverses the push of H′ₘₙ every half period, so a distant level barely fills. The two-level result makes this quantitative: a constant coupling V to a level ħω away transfers at most 4|V|²/(ħ²ω² + 4|V|²), about (2|V|/ħω)². For hydrogen 1s in 1.0 × 10⁷ V m⁻¹, 2p receives at most 6.0 × 10⁻⁹. Truncation compares each coupling with its own splitting, not with the energies, which is why hydrogen's ladder of scales matters: gross structure at α²mc², fine structure at α⁴mc², the Lamb shift at α⁵mc² ln(1/α). A field that leaves 1s alone mixes 2s₁/₂ strongly with 2p₁/₂, only 1057.8 MHz away, as the Hard example shows. Three warnings. A drive at ω turns the phases into e(i(ωₘₙ ± ω)t), so a far level becomes near when the drive is tuned to it; that is what makes a two-level atom in a laser a good model. A level reached only through another enters at second order, the last topic of this unit. And H′ must be prescribed: a quantised field, and the decay it causes, lie outside these equations.
Change one variable at a time
Make the relationship visible.
Set B₁ to zero and nothing moves: with H′ = 0 every cₙ is constant. Raise B₀ at fixed B₁ and the ceiling falls roughly as (B₁/B₀)², which is why a distant level can be dropped. Drag τ and |c↓|² freezes wherever it was; only then is it an energy probability.
SPLITTING f₀ = γB₀/2π2.129 kHz
CEILING OF |c↓|²0.265
|c↓|² FROZEN AFTER τ0.137
UPPER STATE OF H₀ + H′0.071
Live interpretationSPLITTING f₀ = γB₀/2π: 2.129 kHz. CEILING OF |c↓|²: 0.265. |c↓|² FROZEN AFTER τ: 0.137. UPPER STATE OF H₀ + H′: 0.071
Catch the common trap
Explain before calculating.
A two-level system starts in |1⟩, the lower eigenstate of H₀. A constant coupling H′ is switched on at t = 0, and the exact coefficient equations give |c₂(t₁)|² = 0.30 at a time t₁ while H′ is still on. Which statement about that 0.30 is correct?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA proton in B₀ = 50.0 μT along z starts in |↑⟩. At t = 0 a coil adds B₁ = 25.0 μT along x and ΔB = 2.00 μT along z. With H₀ = −γB₀Sz, H′ = −γ(ΔB Sz + B₁Sₓ) and γ = 2.675 × 10⁸ rad s⁻¹ T⁻¹, write the two coefficient equations with numerical rates, check their units, and say which terms can move probability.
- Levels of H₀: E↑ = −ħω₀/2 and E↓ = +ħω₀/2, with ω₀ = γB₀ = 2.675 × 10⁸ × 5.00 × 10⁻⁵ = 1.338 × 10⁴ rad s⁻¹. So ω↓↑ = +ω₀ and ω↑↓ = −ω₀.
- Matrix elements, with Sz = (ħ/2)σz and Sₓ = (ħ/2)σₓ from Quantum Mechanics I: H′↑↑ = −ħγΔB/2 and H′↓↓ = +ħγΔB/2 on the diagonal, H′↑↓ = H′↓↑ = −ħγB₁/2 off it. As rates, γΔB/2 = 268 s⁻¹ and γB₁/2 = 3.34 × 10³ s⁻¹.
- Divide iħ dcₘ/dt = Σₙ H′ₘₙe(iωₘₙt)cₙ by iħ, using −1/i = i: dc↑/dt = i(268 s⁻¹)c↑ + i(3.34 × 10³ s⁻¹)e(−iω₀t)c↓, and dc↓/dt = i(3.34 × 10³ s⁻¹)e(iω₀t)c↑ − i(268 s⁻¹)c↓.
- Units: γB is (rad s⁻¹ T⁻¹)(T) = s⁻¹, matching dc/dt for a dimensionless c, and ω₀t is a pure angle. The off-diagonal entries −(ħγB₁/2)e(∓iω₀t) are complex conjugates of each other, so the matrix is Hermitian and |c↑|² + |c↓|² stays 1.
- Reading: the 268 s⁻¹ terms only turn phases, widening the splitting as if B₀ were 52.0 μT. The 3.34 × 10³ s⁻¹ terms move probability, but their factor e(±iω₀t) reverses the push at a rate four times the coupling rate, so only part of the population can move.
Answerdc↑/dt = i(268 s⁻¹)c↑ + i(3.34 × 10³ s⁻¹)e(−iω₀t)c↓ and dc↓/dt = i(3.34 × 10³ s⁻¹)e(iω₀t)c↑ − i(268 s⁻¹)c↓, with ω₀ = 1.338 × 10⁴ rad s⁻¹; only the B₁ terms move probability.
MediumThe proton of the figure, in B₀ = 50.0 μT and starting in |↑⟩, feels B₁ = 20.0 μT along x from t = 0 until τ = 0.150 ms (γ = 2.675 × 10⁸ rad s⁻¹ T⁻¹). Solve the coefficient equations exactly by eliminating c↑. Find the largest |c↓|², when it is first reached, and the probability that an energy measurement after τ finds the upper level, and compare that with the upper energy state of H₀ + H′.
- With ω₀ = γB₀ = 1.3375 × 10⁴ rad s⁻¹ and ω₁ = γB₁ = 5.350 × 10³ rad s⁻¹, the equations are dc↑/dt = (iω₁/2)e(−iω₀t)c↓ and dc↓/dt = (iω₁/2)e(iω₀t)c↑.
- The second gives c↑ = (2/iω₁)e(−iω₀t) dc↓/dt. Differentiating and substituting into the first: c̈↓ − iω₀ċ↓ + (ω₁²/4)c↓ = 0. Putting c↓ = e(iω₀t/2)f leaves f̈ + (Ω/2)²f = 0, with Ω = √(ω₀² + ω₁²) = 1.4405 × 10⁴ rad s⁻¹.
- Initial values c↓(0) = 0 and ċ↓(0) = iω₁/2 give f = (iω₁/Ω)sin(Ωt/2), so |c↓|² = (ω₁²/Ω²)sin²(Ωt/2). Checks: dimensionless; zero for B₁ = 0; sin²(ω₁t/2), a full flip, for B₀ = 0.
- Ceiling ω₁²/Ω² = B₁²/(B₀² + B₁²) = 400/2900 = 0.138, first reached at t = π/Ω = 0.218 ms. At τ: Ωτ/2 = 1.0804 rad and sin² = 0.7782, so |c↓(τ)|² = 0.138 × 0.7782 = 0.107.
- After τ, H′ = 0 and every dc/dt vanishes, so E↓ = +ħω₀/2 is found with probability 0.107 at any later time. While B₁ was on, the upper state of H₀ + H′, spin against a total field tilted θ = 21.8° from z, held the constant probability sin²(θ/2) = (1 − B₀/|B|)/2 = 0.0358.
Answer|c↓|² = (B₁²/|B|²)sin²(Ωt/2) with Ω = 1.44 × 10⁴ rad s⁻¹; ceiling 0.138, first reached at 0.218 ms; after switch-off P(E↓) = 0.107, against a constant 0.0358 for the upper state of H₀ + H′ while B₁ was on.
HardMetastable hydrogen in 2s₁/₂ (mⱼ = ½) enters a field E = 20.0 kV m⁻¹ along z, switched on abruptly. H₀ includes fine structure and the Lamb shift: 2p₁/₂ lies 1057.8 MHz below 2s₁/₂ and 2p₃/₂ lies 9911 MHz above it, and the field couples 2s₁/₂ to them with |H′| = √3 ea₀E and √6 ea₀E. Decide whether two levels suffice, find the largest 2p₁/₂ population, and say what the model leaves out.
- Couplings: √3 ea₀E = 1.732 × 5.292 × 10⁻¹¹ m × 2.00 × 10⁴ V m⁻¹ = 1.833 μeV, or h × 443.2 MHz, and √6 ea₀E = 2.592 μeV, or h × 626.8 MHz. Both follow from Unit 2's radial functions and Unit 5's Clebsch-Gordan coefficients; 1s is not coupled at all, since ⟨1s|z|2s⟩ = 0 by parity.
- Compare 2|H′| with each splitting: 2 × 443.2/1057.8 = 0.838 for 2p₁/₂ and 2 × 626.8/9911 = 0.126 for 2p₃/₂. For n = 3, 457 THz away with ⟨310|z|200⟩ = 1.77a₀, the ratio is 2.0 × 10⁻⁶. Taken alone, the ceilings 4|V|²/(ħ²Ω²) are 0.413, 0.0157 and 3.9 × 10⁻¹².
- Keep 2s₁/₂ and 2p₁/₂. The two-level solution gives P(2p₁/₂) = 0.413 sin²(Ωt/2) with Ω/2π = √(1057.8² + 886.5²) MHz = 1380 MHz, first peaking at t = π/Ω = 0.362 ns.
- Test the truncation by keeping 2p₃/₂: integrating the three coefficient equations numerically gives a first peak of 0.428 at 0.371 ns, while 2p₃/₂ reaches at most 0.0158. Its coupling pushes 2s down by 626.8²/9911 MHz = 39.6 MHz, narrowing the Lamb gap by 4%, so the two-level answer is about 4% low.
- Orders in α: the n = 3 gap is (5/72)α²mc² = 1.89 eV, derived in Unit 2; fine structure is α⁴mc²/32 = 10.95 GHz, computed in Unit 8 with the Thomas factor quoted from relativistic quantum mechanics, and measured at 10.97 GHz; the Lamb gap is of order α⁵mc² ln(1/α), quoted from quantum electrodynamics and measured at 1057.8 MHz. The smallest splitting picks the level that matters.
- Left out: 2p decays with a measured lifetime of 1.60 ns, about two oscillation periods, a spontaneous decay that no prescribed H′(t) produces; Unit 13 obtains its rate from Einstein's detailed-balance argument and quotes the quantum-electrodynamic derivation. Hyperfine structure splits 2s by 177.6 MHz, 17% of the Lamb gap; and a real field rises in a finite time rather than as a step.
AnswerTwo levels suffice to about 4%: P(2p₁/₂) first peaks at 0.413 after 0.362 ns (0.428 at 0.371 ns with 2p₃/₂ kept, which itself stays below 0.016); n = 3 stays below 10⁻¹¹. Decay and hyperfine structure are left out.