advanced32 min studyLesson 477 in Physics

Derive decay into a continuum of vacuum modes

Connect dipole coupling and mode density to spontaneous emission while delimiting exponential and Markov approximations.

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What you’ll understand

  • Derive the frequency-cubed scaling of a free-space dipole decay rate.
  • Distinguish survival amplitude from excited-population decay.
  • State the weak-coupling continuum assumptions and their short- and long-time limits.

An excited atom can emit into initially empty field modes because the atom–field interaction connects its excitation to a ground-state atom plus a photon. In a continuum, those amplitudes spread over many frequencies and directions, producing approximately irreversible decay on ordinary lifetime scales. A single lossless cavity instead permits coherent return. We will identify how the free-space density of modes and dipole matrix element set the rate, and why an exponential decay law is a controlled approximation rather than an exact statement at every time.

Vacuum supplies accessible final states

Begin with the atom excited and the electromagnetic field in vacuum. In the one-excitation sector the joint state contains a survival amplitude for excited atom plus vacuum and an amplitude for each ground-atom one-photon mode. The rotating-wave dipole interaction couples these components. No classical incident amplitude is needed: photon creation has a nonzero matrix element from vacuum. The coupling to a given mode depends on its field normalization and on the dipole projection onto its polarization.

For free space, count modes in a large quantization volume and sum over two transverse polarizations. The inverse-volume factor in the squared coupling cancels the volume in mode density, so the observable decay rate cannot depend on an arbitrary normalization box. The mode density contributes frequency squared and the field-per-photon coupling contributes another frequency, giving a frequency-cubed rate. The angular and polarization integral produces the stated SI coefficient for an isolated electric-dipole transition.

∣Ψ(t)⟩=ce(t)∣e,0⟩+∑k,λckλ(t)∣g,1kλ⟩,Γ=ω03∣dge∣23πϵ0ℏc3|\Psi(t)\rangle=c_e(t)|e,0\rangle+\sum_{\mathbf k,\lambda}c_{\mathbf k\lambda}(t)|g,1_{\mathbf k\lambda}\rangle,\qquad \Gamma=\frac{\omega_0^3|\mathbf d_{ge}|^2}{3\pi\epsilon_0\hbar c^3}
ω₀ is transition angular frequency in s⁻¹, dge is electric dipole matrix element in C·m, and Γ is population decay rate in s⁻¹. The free-space formula assumes an isolated nonrelativistic electric-dipole transition in vacuum, with appropriate sublevel sums if needed.↔ Scroll wide equations horizontally.

A short bath memory gives exponential decay

Eliminating the photon amplitudes produces a memory integral for the atomic survival amplitude. A broad smooth reservoir spectrum and weak coupling let that amplitude vary slowly during the reservoir correlation time. Replacing its past value by its present slowly varying value gives the Markov approximation. Extending the relevant frequency integration near resonance then gives an amplitude damping term and a transition-frequency shift. The resulting exponential is useful when the bath memory is much shorter than the lifetime.

The survival amplitude decays at half the population rate, so probability decays with Gamma while amplitude has Gamma divided by two in its exponent. A coherent phase also evolves with the shifted transition frequency. This survival amplitude is a component of the joint atom–field state, not automatically the atom’s off-diagonal density-matrix element. An initially excited atom has zero atomic optical coherence, even though its excitation later decays through a nonzero survival amplitude.

ce(t)≃ce(0)e−(Γ/2+iδω)t,Pe(t)≃Pe(0)e−Γt,ρ˙=−iℏ[Heff,ρ]+ΓD[σ−]ρc_e(t)\simeq c_e(0)e^{-(\Gamma/2+i\delta\omega)t},\qquad P_e(t)\simeq P_e(0)e^{-\Gamma t},\qquad \dot\rho=-\frac{i}{\hbar}[H_{\mathrm{eff}},\rho]+\Gamma\mathcal D[\sigma_-]\rho
δω is the environment-induced frequency shift in s⁻¹ in a rotating frame. D[L]ρ=LρL†−(L†Lρ+ρL†L)/2. H_eff is the Hermitian atomic Hamiltonian including the frequency shift, in joules. The reduced Markov equation assumes a vacuum bath, weak coupling, and suitable secular/rotating-wave approximations.↔ Scroll wide equations horizontally.

A lifetime law has a domain

At sufficiently short times, unitary evolution of a state with finite energy variance gives a quadratic survival-probability change, rather than an exact exponential with a nonzero initial slope. At extremely late times, spectral thresholds can generate nonexponential tails. Structured photonic environments, band edges, strong atom–cavity coupling, and delayed feedback can make bath memory important even on experimentally relevant scales. These cases require the reservoir spectrum or explicit field modes rather than a constant free-space decay rate.

A branching transition adds partial population decay rates, and collection efficiency determines what fraction of emitted photons is recorded. A reduced count rate alone does not demonstrate a longer intrinsic lifetime. For independent identically prepared atoms, the number still excited at a specified time is binomial with survival probability given by the assumed lifetime law. The probability lab matches that ensemble count, not a coherent atomic trajectory, a bath memory integral, or the arrival times of emitted photons.

Γtot=∑jΓj,τ=Γtot−1,Bj=ΓjΓtot,Pe(t)=1−(ΔH)2ℏ2t2+⋯(t→0)\Gamma_{\mathrm{tot}}=\sum_j\Gamma_j,\qquad \tau=\Gamma_{\mathrm{tot}}^{-1},\qquad B_j=\frac{\Gamma_j}{\Gamma_{\mathrm{tot}}},\qquad P_e(t)=1-\frac{(\Delta H)^2}{\hbar^2}t^2+\cdots\quad(t\to0)
Γj are partial population rates in s⁻¹, τ is lifetime in seconds, and Bj branching probabilities. The short-time expression assumes an initially excited joint state with finite total-Hamiltonian variance; it is not the Markov exponential expanded to first order.↔ Scroll wide equations horizontally.

Convert half-life into amplitude and population

An initially excited atom has an approximately exponential population half-life of 10 ns in a broad weakly coupled vacuum reservoir. Find Γ, population after 20 ns, and the magnitude of its joint excited-vacuum survival amplitude at that time.

  1. Population half-life obeys exp(−Γ×10 ns)=1/2. Therefore Γ=ln2/(10 ns)≈6.93×10⁷ s⁻¹.

  2. Twenty nanoseconds is two population half-lives, so the remaining excited probability is one quarter. The atom has emitted with probability three quarters if this closed decay has no other channels.

  3. The survival amplitude magnitude is the square root of the remaining probability, equal to one half. Its complex phase depends on the chosen rotating frame and frequency shift; this calculation concerns its magnitude.

The result

Γ≈6.93×10⁷ s⁻¹, excited probability is 1/4 after 20 ns, and survival-amplitude magnitude is 1/2. Population and amplitude have different exponential rates.

Explore the model

Interactive lab

Interactive lab · probability

Count surviving independent atoms

For fresh runs Reset and set BOTH p=0.5, trials n=8; then Reset and set p=0.25, n=8. Interpret p as survival probability at two specified times. Predict mean survivor counts 4 and 2 and variances 2 and 1.5. This exact binomial instrument contains neither emission times nor a vacuum field and assumes independent atoms.

Count surviving independent atomsThese are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Current readings: Mean · np: 5; Variance · np(1−p): 2.5; Total probability: 1. All parameters are adjustable with the labeled controls below.03581000.070.150.220.3Successes kP(X = k)k = 0 · P = 9.77e-4
Exact probabilityHighlighted outcomeMean np
Mean · np
5
Variance · np(1−p)
2.5
Total probability
1
t = 0 s

These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forAn exponential single-atom survival model can imply binomial ensemble counts without making the lab a decay simulator.

Check your understanding

If the excited population decays as exp(−Γt), what is the magnitude of the excited-vacuum survival amplitude in the same Markov model?

Try it yourself

An excited state decays through two channels with partial rates 3.0×10⁶ and 1.0×10⁶ s⁻¹. Find total lifetime and the probability of eventually recording a photon from the first channel if its collection-and-detection efficiency is 0.20.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Add population rates before computing lifetime; multiply the channel branching probability by its efficiency.

Compare with the solution

The total rate is 4.0×10⁶ s⁻¹, giving lifetime 2.5×10⁻⁷ s or 0.25 μs. The first-channel branching probability is 3/4, so eventual recording probability is (3/4)×0.20=0.15. This assumes one initial excitation, no re-excitation, independent detection loss, and no other decay channels.

Take it with you

  • Free-space spontaneous decay combines vacuum mode density with dipole coupling.
  • Amplitude decays at Γ/2 while population decays at Γ.
  • Exponential Markov decay excludes important short-time, late-time, and structured-reservoir effects.

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