University Quantum Mechanics II · Time-Dependent Perturbation Theory · 12.7

A level decaying into a quasi-continuum, solved numerically

The golden rule turned a sum over final states into an integral and leaned on first-order theory. Here nothing is approximated: one level, a ladder of band levels, one matrix. The computer hands back the exponential at exactly the golden-rule rate, then shows you the date it expires, and why.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Quantum Mechanics I solved a level coupled to one partner and found oscillation, never decay; the golden rule found decay by letting the level talk to a continuum. This topic stages the step between them with nothing left out. Put one level |0⟩ at E₀ and N levels spaced 1/ρ apart, couple it to each with the same V, and the Hamiltonian is an (N + 1) × (N + 1) matrix a laptop diagonalises in milliseconds.

Eliminating the band amplitudes exactly leaves a memory kernel, |V|²/ħ² times the sum over k of e^{(−i(Eₖ − E₀)s/ħ)}, and for an equally spaced ladder that sum is a comb of delta functions hρ apart. The tooth at s = 0 gives dc₀/dt = −(Γ/2)c₀ with Γ = 2π|V|²ρ/ħ: the golden-rule rate, with no assumption that c₀ stays near 1. The tooth at s = hρ is the leaked amplitude coming back, because band phases that drift apart at rates spaced 1/(ħρ) realign after hρ.

So a quasi-continuum decays exponentially for exactly as long as it cannot be told from a true one, and the true continuum is the limit ρ → ∞ at fixed |V|²ρ, where the echo never arrives. The model stops describing decay when fewer than about one level lies inside ħΓ, so the echo returns before the decay is done; when the band is not wide compared with ħΓ, so the level feels its edges or, in the extreme, Rabi-oscillates with one bright combination; and when V or ρ vary across ħΓ, which the flat band assumes away.

Simple definition: A quasi-continuum is a band of discrete levels spaced closely enough that a level coupled to them decays exponentially at the golden-rule rate Γ = 2π|V|²ρ/ħ for as long as the spacing cannot be resolved: up to the recurrence time hρ.
Example: Couple a level with |V| = 0.080 µeV to 251 levels spaced 0.10 µeV apart (ρ = 10 µeV⁻¹): ħΓ = 0.402 µeV, the lifetime 1/Γ is 1.64 ns, and the band rephases at hρ = 41.4 ns, about 25 lifetimes later.

The quasi-continuum HamiltonianH = E₀|0⟩⟨0| + Σₖ Eₖ|k⟩⟨k| + Σₖ (V|k⟩⟨0| + V*|0⟩⟨k|), with Eₖ = E₀ + (k + f)/ρ

An (N + 1) × (N + 1) matrix with nothing small in it. Integrate iħ dc/dt = Hc or diagonalise H: either way the answer is exact for this model.

V coupling (J); ρ band levels per unit energy (J⁻¹); f offset of E₀ from the ladder, in spacings; k runs over N levels

Exact memory equation for the leveldc₀/dt = −(|V|²/ħ²) ∫₀ᵗ Σₖ e^{(−i(Eₖ − E₀)(t − t′)/ħ)} c₀(t′) dt′

The kernel is the band's own phases. For an equally spaced ladder it is a comb of delta functions hρ apart, and the comb decides everything else.

c₀ interaction-picture amplitude of |0⟩, dimensionless; the band amplitudes have been integrated out exactly

Decay rate from the tooth at s = 0Γ = 2π|V|²ρ/ħ

Up to t = hρ the infinite ladder obeys dc₀/dt = −(Γ/2)c₀ exactly, so the survival probability is e^(−Γt) right up to the first recurrence.

Γ in s⁻¹: J² × J⁻¹ ÷ (J s). Found in 12.06 at first order; derived here with no assumption that c₀ ≈ 1

Recurrence timeT_rec = hρ = 2πħρ

Band phases (Eₖ − E₀)t/ħ differ by multiples of 2πt/(hρ), so all realign at t = hρ and the amplitude that leaked into the band returns to |0⟩.

T_rec in s; with h = 4.1357 µeV ns, a spacing 1/ρ = 0.10 µeV rephases after 41.4 ns

Lifetimes that fit before the echoΓT_rec = 2π ħΓρ = 4π²|V|²ρ²

With ħΓρ = 1 the level is down to e^(−2π) = 0.0019 when the band rephases; with ħΓρ = 0.1 it still holds 53 per cent, so the echo arrives before even half the population has left.

dimensionless; ħΓρ is the number of band levels lying inside the width ħΓ

Weight of |0⟩ in each eigenstate (infinite ladder)|⟨0|j⟩|² = |V|² / [(Eⱼ − E₀)² + |V|² + (ħΓ/2)²]

A Lorentzian of full width √((ħΓ)² + 4|V|²), which tends to ħΓ once many levels lie inside ħΓ, sampled once per spacing. Its Fourier transform is the exponential; the sampling is what brings the echo.

Eⱼ eigenvalue of H (J), one in each gap between band levels; the weights are dimensionless and sum to 1

Build the matrix, and keep every term

Put the discrete level at E₀ = 0 and N band levels at Eₖ = (k + f)/ρ, spread symmetrically about it, each coupled with the same real V. Written for Schrödinger-picture amplitudes, the coefficient equations of topic 12.01 become iħ dc₀/dt = V Σₖ cₖ and iħ dcₖ/dt = Eₖcₖ + Vc₀: N + 1 coupled equations with no small parameter. Two routes solve them. Integrate from c₀ = 1 with a fourth-order Runge–Kutta step, using Σ|c|² = 1 as the running check; or diagonalise H once and write P(t) = |Σⱼ |⟨0|j⟩|² e^{(−iEⱼt/ħ)}|², exact at every t for the cost of one eigenproblem. For the base example, 251 levels 0.10 µeV apart with |V| = 0.080 µeV, a 2 ps step holds the norm to 2 × 10⁻¹⁰ over 5 ns, and both routes give P(5 ns) = 0.0466682. Whatever the computer returns is a property of the model, not of a series; the only idealisations are those written into H: a flat band, one coupling, finite N.

Integrate out the band: a comb of delta functions

In the interaction picture each band amplitude is cₖ(t) = −(iV/ħ) ∫₀ᵗ e^(iωₖt′) c₀(t′) dt′, with ωₖ = (Eₖ − E₀)/ħ. Substituting gives dc₀/dt = −∫₀ᵗ K(t − t′) c₀(t′) dt′ with K(s) = (|V|²/ħ²) Σₖ e^(−iωₖs), still exact. For an infinite ladder ωₖ = (k + f)/(ħρ), and a Fourier series with equal coefficients is a comb: Σₖ e^(−ikx) = 2π Σₙ δ(x − 2πn). So K(s) = Γ Σₙ e^(−2πifn) δ(s − nhρ), with Γ = 2π|V|²ρ/ħ. The tooth at s = 0 sits at the end of the integration range and counts half, so for all t up to hρ the equation is dc₀/dt = −(Γ/2)c₀ and P(t) = e^(−Γt) exactly. That is the golden-rule rate, derived without ever assuming c₀ ≈ 1. The base-example matrix, a finite band, bears it out: P(5.0 ns) = 0.0467 against e^(−Γt) = 0.0471.

Diagonalise: the level's weight is a sampled Lorentzian

The eigenbasis tells the same story in energy. The components of H|j⟩ = Eⱼ|j⟩ give the secular equation Eⱼ − E₀ = Σₖ |V|²/(Eⱼ − Eₖ): one root in each gap between band levels, plus one below the band and one above. Normalising the eigenvector gives |⟨0|j⟩|² = [1 + Σₖ |V|²/(Eⱼ − Eₖ)²]⁻¹, and for the infinite ladder the table sum Σₖ 1/(x − k)² = π²/sin²(πx) turns this into |V|²/[(Eⱼ − E₀)² + |V|² + (ħΓ/2)²]: a Lorentzian of full width √((ħΓ)² + 4|V|²), which tends to ħΓ once many levels lie inside ħΓ, sampled about once per spacing. In the base example that width is 0.433 µeV against ħΓ = 0.402 µeV, the central weights are 0.1317 and 0.1004, against 0.1314 and 0.0997 from the formula, and the four eigenstates within ħΓ/2 of E₀ hold 46 per cent of the weight, near the 48 per cent a continuous Lorentzian of the same width holds there. In the dense limit the Fourier transform of that Lorentzian is e^(−Γ|t|/2), the decaying amplitude; sampling it at points about 1/ρ apart makes the transform come back after hρ.

The echo at hρ, and the shape it takes

The tooth at s = hρ is amplitude that entered the band at t′ returning at t′ + hρ: the band phases (Eₖ − E₀)t/ħ are spaced 2πt/(hρ) apart, so at t = hρ they all agree again and the band interferes back onto |0⟩. Between hρ and 2hρ the memory equation reads dc₀/dt = −(Γ/2)c₀ − Γe^(−2πif) c₀(t − hρ); inserting the first-window solution c₀(t − hρ) = e^(−Γτ/2) on the right, it integrates to c₀ = e^(−Γt/2) − e^(−2πif) Γτ e^(−Γτ/2), with τ = t − hρ. When ΓT_rec ≫ 1 the first term has died and P ≈ (Γτ)² e^(−Γτ), peaking at τ = 2/Γ with 4/e² = 0.541. The 252 × 252 base matrix peaks at 0.553 at τ = 3.23 ns, against 3.27 ns predicted. A physical version: an emitter at the centre of a closed one-dimensional cavity of length L couples only to every other mode, spaced hc/L, so hρ = L/c, the light's trip to a mirror and back. The true continuum is ρ → ∞ at fixed |V|²ρ: Γ is unchanged and the echo never comes.

Turn each assumption into a parameter and break it

Fewer levels inside ħΓ: since P = e^(−Γt) holds exactly until hρ, the damage is that only ΓT_rec = 2πħΓρ lifetimes fit in the window. With |V| = 0.020 µeV and spacing 0.10 µeV, ħΓρ = 0.25; the level still holds 21 per cent when the band rephases at 41 ns and is back at 0.81 by 100 ns. A narrower band: keep |V| = 0.080 µeV and spacing 0.10 µeV but cut N. With 251 levels, W = 62ħΓ, the matrix decays 1 per cent faster than Γ, an edge effect of order ħΓ/W; with 11 levels, W = 2.7ħΓ, the matrix gives P(2 ns) = 0.51 against e^(−Γt) = 0.29, then 0.002 at 6 ns and 0.011 at 8 ns. When W ≪ √N|V| the level sees only the symmetric state Σₖ|k⟩/√N, coupled with strength √N|V|, and the two-state result returns: P = cos²(√N|V|t/ħ). For N = 5 at spacing 0.010 µeV the level empties at 5.78 ns and refills, where the golden rule would claim a 0.16 ns lifetime.

Why an atom's decay passes every test

Each result above can be checked three ways. Dimensions: Γ = 2π|V|²ρ/ħ has units J² × J⁻¹ ÷ (J s) = s⁻¹, and hρ has J s × J⁻¹ = s. Limits: V → 0 leaves P = 1; ρ → ∞ at fixed |V|²ρ keeps Γ and sends the echo to infinity; a band squeezed to zero width gives back the Rabi oscillation of Quantum Mechanics I. Order in α and measurement: the model contains no α until the band is identified with photon modes, and then ħΓ comes out at order α³ times the atomic energy scale (derived in Unit 13, quoted here). For hydrogen 2p, ħΓ/Ry = 2(2/3)⁸α³ = 3.0 × 10⁻⁸, a lifetime of 1.60 ns, as measured. The photon continuum changes only on the scale of the 10.2 eV transition energy, 2.5 × 10⁷ times ħΓ = 0.41 µeV, and in open space it has no spacing at all, so both conditions hold by factors of millions. That is why a measured Lyman-α decay curve is a clean exponential.

Change one variable at a time

Make the relationship visible.

Interactive model

Coupling |V| to each band level0.25 µeV

Band level spacing 1/ρ0.5 µeV

Offset f of E₀ from a band level0.0 spacings

Default: the solid curve rides the golden-rule line to the floor, then climbs back to about one half 1.7 ns after hρ = 8.3 ns. At |V| = 0.15 µeV, slide f from 0 to 0.5: nothing changes before hρ, but the notch just after it, where the echo cancels the remainder, vanishes.

WIDTH ħΓ = 2π|V|²ρ0.785 µeV

LIFETIME 1/Γ0.84 ns

RECURRENCE hρ8.27 ns

LEVELS INSIDE ħΓ1.57

Live interpretationWIDTH ħΓ = 2π|V|²ρ: 0.785 µeV. LIFETIME 1/Γ: 0.84 ns. RECURRENCE hρ: 8.27 ns. LEVELS INSIDE ħΓ: 1.57

Catch the common trap

Explain before calculating.

Misconception check

“The golden rule is first-order perturbation theory, so the exponential can only be trusted while nearly all the population is still in the initial level.” The numerical solution refutes this directly: with no approximation anywhere, the base-example level follows e^(−Γt) down to a ten-thousandth of its population (P(15 ns) = 9.7 × 10⁻⁵ against 1.05 × 10⁻⁴), far past the point where c₀ ≈ 1 has failed. First-order theory supplies only the initial slope. What keeps the decay exponential is that the band forgets: its memory kernel is a delta function at s = 0, so the rate at which amplitude leaves |0⟩ depends on c₀ now, whatever its size. The assumptions that matter are about the band, dense, wide and flat, and the times at which the exponential fails are set by them: hρ, when the memory returns, and the short times before the band's width is resolved. Depletion is not one of them.

A level is coupled with |V| = 0.050 µeV to each of 1001 band levels spaced 0.020 µeV apart, and the 1002 × 1002 Hamiltonian is diagonalised exactly. The survival probability falls exponentially, close to the golden-rule lifetime of 0.84 ns. When does it first come back above one half?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA level is coupled with |V| = 0.050 µeV to each of 401 band levels spaced 0.025 µeV apart and centred on it. Find ħΓ, the lifetime, the recurrence time and the number of band levels inside ħΓ, and decide whether the numerical solution will show a clean exponential window.

Density of states ρ = 1/(0.025 µeV) = 40 µeV⁻¹. Width ħΓ = 2π|V|²ρ = 2π × (0.050 µeV)² × 40 µeV⁻¹ = 0.628 µeV; the units µeV² × µeV⁻¹ = µeV are those of a width.
Lifetime 1/Γ = ħ/(ħΓ) = 0.6582 µeV ns ÷ 0.628 µeV = 1.05 ns.
Recurrence hρ = 4.1357 µeV ns × 40 µeV⁻¹ = 165 ns, so ΓT_rec = 165.4 ns ÷ 1.048 ns = 158 lifetimes fit before the band rephases.
Density test: ħΓρ = 0.628 µeV × 40 µeV⁻¹ = 25 levels inside the width, and 2π × 25.1 = 158 agrees with step 3. The level is down to e⁽⁻¹⁵⁸⁾ long before the echo.
Width test: W = 401 × 0.025 µeV = 10.0 µeV = 16ħΓ, so edge corrections of order ħΓ/W, a few per cent in the rate, are expected. The diagonalised 402 × 402 matrix gives P(5/Γ) = 6.1 × 10⁻³ against e⁻⁵ = 6.7 × 10⁻³.

AnswerħΓ = 0.628 µeV, 1/Γ = 1.05 ns, hρ = 165 ns and 25 levels inside ħΓ: a clean exponential window of about 158 lifetimes, with the rate good to a few per cent because the band is only 16ħΓ wide.

MediumA level sits in a ladder of levels spaced 0.20 µeV apart and is coupled to each with |V| = 0.060 µeV. Using the exact memory equation, find the survival probability just before the first recurrence and at a time 2/Γ after it, first with E₀ exactly on a band level (f = 0), then midway between two (f = ½).

ρ = 1/(0.20 µeV) = 5.0 µeV⁻¹, so ħΓ = 2π × (0.060 µeV)² × 5.0 µeV⁻¹ = 0.113 µeV, Γ = 0.113 µeV ÷ 0.6582 µeV ns = 0.172 ns⁻¹ and 1/Γ = 5.82 ns. The recurrence is at hρ = 4.1357 µeV ns × 5.0 µeV⁻¹ = 20.7 ns.
Before hρ only the s = 0 tooth of the kernel acts, so P = e^(−Γt) exactly whatever f is: P just before hρ is e^(−ΓT_rec) = e^(−3.55) = 0.0286. Only ħΓρ = 0.57 levels lie inside ħΓ, so an amplitude e^{(−ΓT_rec/2)} = 0.169 is still in |0⟩ when the echo arrives.
After hρ the second tooth feeds the delayed amplitude back: c₀ = e^(−Γt/2) − e^(−2πif) Γτ e^(−Γτ/2), with τ = t − hρ, valid until 2hρ = 41.4 ns.
At τ = 2/Γ = 11.6 ns, so t = 32.3 ns, the remainder is e^(−Γt/2) = e^(−2.78) = 0.0623 and the echo is Γτ e^(−Γτ/2) = 2/e = 0.736.
f = 0: e^(−2πif) = 1 and the two subtract, c₀ = −0.674, P = 0.454. f = ½: e^(−iπ) = −1 and they add, c₀ = 0.798, P = 0.637. A 4001-level diagonalisation gives 0.4537 and 0.6369.

AnswerP = 0.0286 just before hρ = 20.7 ns for either f; at t = 32.3 ns, P = 0.454 with E₀ on a band level and 0.637 with it midway, against 0.541 if the leftover amplitude were ignored.

HardBuild a numerical stand-in for the decay of hydrogen 2p, taking the lifetime τ = 1.596 ns (calculated, and matching the measured 1.60 ns). Use a flat band centred on E₀, and require the first recurrence no sooner than 20 lifetimes and a band at least 50ħΓ wide. Find the largest spacing, the coupling |V|, the smallest N and the matrix size; then say what the matrix shows if N is cut to 11 at the same spacing and coupling.

ħΓ = ħ/τ = 0.6582 µeV ns ÷ 1.596 ns = 0.4124 µeV. The recurrence condition hρ ≥ 20τ = 31.92 ns gives 1/ρ ≤ h/(20τ) = 4.1357 µeV ns ÷ 31.92 ns = 0.1296 µeV.
From Γ = 2π|V|²ρ/ħ: |V|² = ħΓ/(2πρ) = 0.4124 µeV × 0.1296 µeV ÷ 2π = 8.50 × 10⁻³ µeV², so |V| = 0.0922 µeV. Check: ħΓρ = 0.4124/0.1296 = 3.18 = 20/(2π), exactly what ΓT_rec = 2πħΓρ = 20 demands.
Width: N/ρ ≥ 50ħΓ requires N ≥ 50 × 3.183 = 159.2, so N = 160 levels and a 161 × 161 matrix, with E₀ midway between the two central levels.
Diagonalising it: P(τ) = 0.373 against e⁻¹ = 0.368, and P(10τ) = 4.10 × 10⁻⁵ against e⁻¹⁰ = 4.54 × 10⁻⁵, a rate about 1 per cent high, an edge correction of order ħΓ/W = 0.02. The echo peaks at 0.556 at 35.0 ns, against 4/e² = 0.541 at hρ + 2τ = 35.1 ns.
Cut to N = 11, still centred, so one level now sits at E₀: W = 1.43 µeV = 3.5ħΓ. Now P(τ) = 0.574, P reaches 1/e only at 2.14 ns, stalls near 0.02 between 5 and 7 ns and falls to zero at 9.4 ns, where the amplitude, real for this symmetric band, changes sign: no exponential window at all.
Order in α, quoted from Unit 13: ħΓ/Ry = 0.4124 µeV ÷ 13.606 eV = 3.03 × 10⁻⁸, matching 2(2/3)⁸α³ = 3.03 × 10⁻⁸. The stand-in reproduces the rate, not the atom: the real photon continuum has ρ → ∞, and its echo never comes.

Answer1/ρ = 0.1296 µeV, |V| = 0.0922 µeV and N = 160, a 161 × 161 matrix, reproduce e^{(−t/1.596 ns)} with the rate good to about 1 per cent and an echo of 0.56 at 35 ns; with 11 levels the decay is not exponential at all.